Binomial Trials And Binomial Distribution

BINOMIAL TRIALS AND BINOMIAL DISTRIBUTION

Consider a random experiment whose outcomes can be classified as success or failure. It means that experiment results in only two outcomes E₁(success) or E₂­ (failure). Further assume that experiment can be repeated several times, probability of success or failure in any trial are p and q (p + q = 1) and don't vary from trial to trial and finally different trials are independent. Such a experiment is called Binomial experiment and trials are said to be binomial trials. For instance tossing of a fair coin several times, each time outcome would be either a success (say occurrence of head) or failure (say occurrence of tail).

A probability distribution representing the binomial trials is said to binomial distribution.

Let us consider a Binomial experiment which has been repeated 'n' times. Let the probability of success and failure in any trial be p and q respectively and we are interested in the probability of occurrence of exactly 'r' successes in these n trials. Now number of ways of choosing 'r' success in 'n' trials = ⁿC_r. Probability of 'r' successes and (n-r) failures is p^r×q^n-r. Thus probability of having exactly r successes = ⁿC_r×p^r×q^n-r

Let 'X' be random variable representing the number of successes, then

P(X = r) = ⁿC_r×p^r×q^n-r (r = 0, 1, 2, L , n)

1 = (p + q)ⁿ = ⁿC₀p⁰qⁿ + ⁿC₁p¹q^{n –1} + ⁿC₂p²q^{n –2} + ……. + ⁿC_rp^rq^{n –r} + ……. + ⁿC_npⁿ

X $\to$ Number of success 0, 1, 2, …….., r, ……….. n

Remarks:

$\bullet$ Probability of utmost 'r' successes in n trials = $\sum _{{}^{r=r}}^r{}^n{C}_r{p}^r{q}^{n-r}$

$\bullet$ Probability of atleast 'r' successes in n trials = $\sum _{r=r}^n{}^n{C}_r{p}^r\cdot q^{n-r}$

$\bullet$ Probability of having 1^st success at the r^th trial = p×q^r-1

Example -: A die is thrown 7 times. What is the chance that an odd number turns up (i) exactly 4 times (ii) at least 4 times.

Sol: Probability of success = $\frac{3}{6}=\frac{1}{2}$ $\Rightarrow$ p = $\frac{1}{2},\text{ q=}\frac{\text{1}}{\text{2}}$

(i) For exactly 4 successes required probability = ⁷C₄. ${\left(\frac{1}{2}\right)}^4.{\left(\frac{1}{2}\right)}^3=\frac{35}{128}$

(ii) For atleast 4 successes required probability

= ⁷C₄ ${\left(\frac{1}{2}\right)}^4{\left(\frac{1}{2}\right)}^3{+}^7{C}_5{\left(\frac{1}{2}\right)}^5.{\left(\frac{1}{2}\right)}^2{+}^7{C}_6{\left(\frac{1}{2}\right)}^6.{\left(\frac{1}{2}\right)}^1{+}^7{C}_7{\left(\frac{1}{2}\right)}^7$

= $\frac{35}{128}+\frac{21}{128}+\frac{7}{128}+\frac{1}{128}=\frac{64}{128}=\frac{1}{2}$